Description and Application of Genetic Algorithm

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Description and Application of Genetic

Algorithm

MIN WANG

Thesis for the Master of Science Degree (two years) in Mathematical Modeling and Simulation

30 credit points (30 ECTS credits) April 2012

Blekinge Institute of Technology School of Engineering

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Abstract

Genetic Algorithm (GA) as a class of Evolutionary Algorithm (EA) is a search algorithm based on the mechanics of natural selection and natural genetics. This dissertation presents the description, solving procedures and application of GA. The definitions of selection, crossover and mutation operators are given in details and an application based on GA in Time Table Problem (TTP) is performed in a new way.

Due to its high capability of overall search, GA is particularly appropriate for solving timetabling and scheduling problems. TTP (Time Table Problem) which belongs to NP-hard problem is a special problem concerning resource management. In this dissertation, a new chromosome coding is designed in order to solve TTP more effectively. And the result presented by MATLAB will converge to a steady condition.

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Acknowledgement

It is a pleasure to express my gratitude to all those who gave me the possibility to complete this thesis.

Firstly, I owe my deepest gratitude to my dear professor Elisabeth Rakus-Andersson. Her kindness, patience, rigorous academic attitude and profound knowledge inspire me. The entire process of this dissertation from topic selection and design plan until completion cannot be accomplished without her meticulous guidance.

Next, I would like to show my sincere appreciation to Mohammad Havaei. He helped me to edit the MATLAB program and put forward many valuable suggestions to my thesis.

Furthermore, I am grateful to professor Nail Ibragimov, Claes Jogréus, Mattias Dahl, Raisa Khamitova, Mattias Eriksson and other professors and managers. During two years of master education, I grasp much precious knowledge.

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Genetic Algorithm (GA) is a class of Evolutionary Algorithm (EA) which generates solutions to optimization problems using techniques inspired by natural evolution, such as inheritance, mutation, selection, and crossover. It is a search algorithm based on the mechanics of natural selection and natural genetics to solve usually mathematical optimization of search algorithm in computational algorithm (Chu & Beasley, 1995).

The idea of GA comes from the natural rule (Davis, 1991). Each organism lives with specific features coded in genes which are stored in chromosomes. They are allowed to transfer their genes to create new generations, and this process is called crossover. During reproduction, mutation possible occurs which expresses in changing one gene in chromosome of offspring. The offspring which can fit to the environment better will be left. GA imitates this hereditary procedure in the mathematical way, which means that after selection, crossover and mutation, the best result whose fitness is highest will be obtained.

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This dissertation is going to introduce this interesting method by applying the genetic algorithm to solve a time table problem (TTP). The author builds up a new model to analyze TTP. This model merges two matrices to create the code of chromosomes. In this way, the methodology to solve this problem becomes more rational. The problem will be simplified as the model only uses binary code with 0 and 1.

2. Background

John Holland firstly put forward GA in 1960s (Wikipedia, 2010), when he worked on the studies of cellular automata with his colleagues and his students at the University of Michigan. GA became popular through his book Adaptation in Natural and Artificial Systems (Holland, 1975). The research of GA was limited to the theoretical part until the First International Conference on Genetic Algorithm, which was held in Pittsburgh, Pennsylvania in the mid-1980s. As the development of computer programs and the demand of practical application grew, GA becomes more popular within practical application.

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been developed rapidly. Most of Fortune 500 companies apply GA to making a time list, data analysis, the future trend forecast, budget, and solving other combinatorial optimization problems.

3. Description of genetic algorithm

3.1 Some definitions

Let us introduce the important definitions cited after the book “Computational Intelligence” written by Leszek Rutkowski (Rutkowski, 2005).

Definition 3.1.1:

Population is a set of individuals of a specified size. Definition 3.1.2:

Individuals of a population are sets of “task parameters” coded in the form of chromosomes.

Definition 3.1.3:

Chromosomes (chains or code sequences) are ordered sequences of genes.

Definition 3.1.4:

Gene (“feature”, ”sign”, ”detector”) is a single element in the genotype. Definition 3.1.5:

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(can be one chromosome). Definition 3.1.6:

Phenotype is a set of values corresponding to a given genotype, which is a decoded structure.

Definition 3.1.7:

Fitness function (“adaptation function”, “evaluation function”) which measures fitness of adaptation of a given individual to the environment. The function allows evaluating the degree of fitness of particular individuals in the population. Basing on this degree we select the individuals that fit best.

Definition 3.1.8:

The new generation (“the offspring generations”) is a newly created population of individuals.

3.2 Solving steps

3.2.1 Flow-process diagram of GA

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Figure 3.2.1-1 Solving procedures of GA

NO

Initiation-selection of the initial population of chromosomes

Stopping criterion

Evaluation of the fitness of chromosomes in the population START Creation of a new population Application of genetic operators

Selection of chromosomes _{of the “best”}Presentation chromosome

STOP

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3.2.2 Initialization

An initial population consists of random selection of a demanded number of individuals. Every individual is represented by chromosomes coded in a determined length of sequence of digits, letters or other showing ways.

Example 3.2.2-1:

Usually, we use binary sequences to create the codes of chromosomes such as

𝑐ℎ₁ = [01010101], 𝑐ℎ₂ = [10010100]

3.2.3 Evaluation of fitness function

The fitness function is used to measure adaptability of chromosomes and usually need to be maximized. The form of the fitness function depends on the real problem. And a suitable transform seems necessary when we face to a minimum optimization problem.

Example 3.2.3-1:

A minimum optimization problem is given as below,

min 𝑓 = 2𝑥, x ≥ 0.

Usually we make a reciprocal transformation to change this minimum problem into a maximum problem, such as

max 𝐹 = 1

1 + 𝑓 =

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Here we use _1+𝑓1 instead of 1_𝑓 in case 𝑓 = 0 . Therefore the maximum function 𝐹 can be treated as the fitness function of this problem.

After determining the fitness function of solved problem, we calculate the evaluation of each chromosome according to the fitness function. The higher the value of the function is, the better result of the solved maximization problem is got.

Example 3.2.3-2:

Calculate the evaluations of the fitness function 𝑓 = 𝑥2_{+ 3 which}

we wish to maximize for two chromosomes 𝑐ℎ₁ = [0100], 𝑐ℎ₂ =

[1010].

Due to 𝑐ℎ₁ = [0100] is a binary sequence, so the chromosome

𝑐ℎ₁ = [0100] is corresponding to

𝑥₁ = 23_{∙ 0 + 2}2_{∙ 1 + 2}1_{∙ 0 + 2}0_{∙ 0 = 4}

Thus the evaluation of this chromosome is

𝑓₁ = 𝑥2_{+ 3 = 4}2_{+ 3 = 19}

The same steps to calculate the evaluation of 𝑐ℎ₂ = [1010]

𝑥₂ = 23_{∙ 1 + 2}2_{∙ 0 + 2}1_{∙ 1 + 2}0_{∙ 0 = 10}

𝑓₂ = 𝑥2_{+ 3 = 10}2_{+ 3 = 103}

We can see that 𝑓₂ = 103 is larger than 𝑓₁ = 19 , so the

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3.2.4 Selection

After calculating the evaluation of every chromosome, we should choose the same number of chromosomes as in the previous population to create the next generation. This selection takes place in accordance with the natural selection rule that the better parents will have better offspring. Parents are chosen according to their fitness. Thus the chromosomes which have highest evaluation should have the most of the chances to be selected to create new offspring. The selection can be done by many methods, such as roulette wheel selection, Boltzmann selection, tournament selection, rank selection, steady state selection and some others (Obitko, 1998).

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Example 3.2.4-1:

Assume that we want to find the maximum value of the function:

𝑦 = 2𝑥, 𝑥 ∈ [0,31]

We start with drawing the initial populations as following numbers from [0, 31]: 16, 21, 6, 30, 14 and 2.

If we use the numbers’ binary codes as their chromosomes, we obtain

𝑐ℎ₁ = [10000], 𝑐ℎ₂ = [10101], 𝑐ℎ₃ = [00110],

𝑐ℎ₄ = [11110], 𝑐ℎ₅ = [01110], 𝑐ℎ₆ = [00010].

According to the fitness function, we get the evaluations of chromosomes as following,

𝑓₁ = 2 ∙ 16 = 32, 𝑓₂ = 2 ∙ 21 = 42, 𝑓₃ = 2 ∙ 6 = 12,

𝑓₄ = 2 ∙ 30 = 60, 𝑓₅ = 2 ∙ 14 = 28, 𝑓₆ = 2 ∙ 2 = 4.

Now we apply roulette wheel method to select chromosomes. The sum of fitness function values of all of chromosomes is

𝐹 = 𝑓₁ + 𝑓₂+ 𝑓₃+ 𝑓₄+ 𝑓₅+ 𝑓₆ = 178

Then we get the sectors of roulette wheel showed in percentage.

𝜗₁ =𝑓1 𝐹 = 32 178 = 0.18, 𝜗2 = 𝑓₂ 𝐹 = 42 178 = 0.23, 𝜗₃ =𝑓3 𝐹 = 12 178 = 0.07, 𝜗4 = 𝑓₄ 𝐹 = 60 178 = 0.34, 𝜗₅ =𝑓5 𝐹 = 28 178 = 0.16, 𝜗6 = 𝑓₆ 𝐹 = 4 178= 0.02.

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If we assign an interval [0, 100) to the whole roulette wheel, the

sector of 𝑐ℎ₁ belongs to the interval [0, 18), as same, the sector of

𝑐ℎ₂ is between [18, 41) and so on. We can see that the chromosome

whose fitness is higher has a larger area of the roulette wheel, and then this chromosome will have more chance to be selected.

Now we choose numbers from interval [0, 100) randomly to help us to select the parents’ chromosomes. Assume that 6 numbers are drawn as following: 89, 14, 19, 73, 47 and 53.

According to Figure 3.2.4-1, we know that number 89 is in the

interval [82, 98) which means that 𝑐ℎ₅ is selected. In the same way,

14 ∈ [0, 18) → 𝑐ℎ₁, 19 ∈ [18, 41) → 𝑐ℎ₂, 73 ∈ [48, 82) → 𝑐ℎ₄,

47 ∈ [41, 48) → 𝑐ℎ₃, 53 ∈ [48, 82) → 𝑐ℎ₄.

We find the following chromosomes have been selected,

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𝑐ℎ₅, 𝑐ℎ₁, 𝑐ℎ₂, 𝑐ℎ₄, 𝑐ℎ₃, 𝑐ℎ₄.

3.2.5 Crossover

In natural world, the genes of children are from both of their parents. In order to imitate this situation, we do crossover of chromosomes of selected parents to generate offspring.

Crossover just happens between two chromosomes (one couple), so the first stage of crossover is arranging the chromosomes of selected parents’ population in pairs which is made randomly.

After that, we should decide the crossover point. If the length of each chromosome is L (i.e., for ch = [10000] , the length of this

chromosome is L=5), we draw a random number 𝑙_𝑘, which belongs

to interval [1, L-1]. Then the genes of parents are exchanged from the

gene on position 𝑙_𝑘. As a result a pair of offspring is created.

Example 3.2.5-1:

If L=8, 𝑙_𝑘 = 5 is drawing, then the crossover happens as following:

In classical GA, crossover is almost always present, thus we give a

pretty high crossover rate 𝑃_{𝐶𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟}, which is generally in the

interval [0.5, 1]. For each pair of parents we randomly draw a value of

A pair of parents A pair of offspring

𝑐ℎ₁ = [10001 𝟎𝟏𝟎]

𝑐ℎ₂ = [01000 𝟏𝟎𝟎]

𝑐ℎ₁′ = [10001 𝟏𝟎𝟎]

𝑐ℎ₂′ = [01000 𝟎𝟏𝟎]

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a crossover probability 𝑃_𝑐 for this pair. If this probability is less than 𝑃_{𝐶𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟} then crossover happens in pair.

3.2.6 Mutation

Different from crossover, mutation occurs quite rarely in GA,

therefore we give a very small mutation rate 𝑃_𝑀, which is often in the

interval [0, 0.1]. Every gene should be given a random mutation

probability 𝑃_𝑚. If the drawn probability is less than or equal to 𝑃_𝑀,

then this gene value should be changed to opposite value (in binary code case, from 0 to 1, or from 1 to 0), which means that mutation occurs.

Example 3.2.6-1:

We perform mutation on chromosome [01101].

Assume that 𝑃_𝑀 = 0.02, the probability of each gene is drawn as

following: 0.34, 0.89, 0.01, 0.50 and 0.78.

We can see that the rate of the third gene is 0.01<𝑃_𝑀 = 0.02, so this

gene should be changed from 1 to 0. As a result, after mutation, this chromosome becomes [01001].

3.2.7 Obtaining new generation

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feasible solutions at all. Therefore we need modify them or compel them to satisfy the conditions. After that we obtain new feasible solutions, so called current population, with the same size of their parents’ population.

3.2.8 Stopping criterion

The criterion for stopping the genetic algorithm depends on the specific situation of the application of GA.

Situation 1:

If we get the best result 𝑐ℎ_𝑖 after i generations, and we are no longer

get better solution in next generations 𝑖 + 1, 𝑖 + 2, 𝑖 + 3 ⋯, then we can stop the algorithm.

Situation 2:

If we have a desired value 𝑓_𝑑of the problem, then the program can be

stopped when we get the approaching result, such as in generation i.

Evaluation 𝑓_𝑖 of the selection 𝑐ℎ_𝑖 is compared to expected value 𝑓_𝑑.

If 𝑓_𝑖 − 𝑓_𝑑 ≤ 𝜀(𝜀 𝑖𝑠 𝑎 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑙𝑖𝑘𝑒 𝜀 = 10−15_{) , then the}

algorithm will be stopped. Situation 3:

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If the stopping criterion is met then the best result is taken, otherwise go back to selection step.

3.2.9 Getting the best result

The stop criterion should be checked when we obtain a new generation. If it is satisfied, then we stop the whole program and output the final answer. Otherwise we will go back to the selection step until we achieve the stop criterion. The best result should be a chromosome which has a highest fitness function value.

4. Application of the genetic algorithm to make a university

courses schedule

Time Table Problem (TTP) as a NP-complete (non-deterministic polynomial-time) problem is a typical portfolio optimization and uncertainty scheduling problem, aiming at solving the resources arrangement between time and space. It has been investigated by many researchers albeit with respect to the timetabling of resources in educational institutions (Cole, 1964) (Wood, 1968). The author creates a new model to deal with TTP by GA.

4.1 TTP-model

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class. In this model, the definition of each element has been given as below.

4.1.1 Time

Time-plot set:T = {𝑡₁, 𝑡₂, ⋯ , 𝑡_𝑎, ⋯ 𝑡_𝐴}

1. We suppose that every class has lectures weekly.

2. Professors and students do not have lectures on Saturday and Sunday. 3. The minimum unit of one class-time is two hours, so we separate every

workday into 4 time-plots: 8:00-10:00, 10:00-12:00, 13:00-15:00 and

15:00-17:00. So from the time-plot set, 𝑡₁means from 8:00 to 10:00 on

Monday, 𝑡₂shows from 10:00 to 12:00 on Monday and so on, and

𝑡_𝐴means the last time-plot from 15:00 to 17:00 on Friday. Obviously, we

get A = T = 5 × 4 = 20.

4.1.2 Room (Place)

Room set: R = {𝑟₁, 𝑟₂, ⋯ , 𝑟_𝑏, ⋯ 𝑟_𝐵}

Every room has its certain seating capacity, we use |𝑟_𝑏 to show the

seating capacity of room 𝑟_𝑏.

Time-plots respond to rooms, their Cartesian product is

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4.1.3 Professor

Professor set: P = {𝑝₁, 𝑝₂, ⋯ , 𝑝_𝑐, ⋯ 𝑝_𝐶}

Professors are knowledgeable, mostly each professor do not only teach

one course. We suppose that professor 1 (𝑝₁) can teach 𝛼₁ courses,

professor 2 (𝑝₂) can teach 𝛼₂ courses, and so on.

4.1.4 Lecture (Course)

Lecture set: L = {𝑙₁, 𝑙₂, ⋯ , 𝑙_𝑑, ⋯ 𝑙_𝐷}

We assume that each course is only taught by one professor.

Let us rank the elements of lecture set based on the order of professors.

i.e., professor 1 teaches courses from 𝑙₁ to 𝑙_𝛼₁, then professor 2

teaches courses between 𝑙_𝛼₁₊₁ and 𝑙_𝛼₁_+𝛼₂ and go on. Then

L =

{𝑙₁, 𝑙₂, ⋯ 𝑙_𝛼₁, 𝑙_1+𝛼₁, ⋯ , 𝑙_𝛼₁_+𝛼₂, 𝑙_1+𝛼₁_+𝛼₂, ⋯ , 𝑙_𝛼₁_+𝛼₂_+𝛼₃, ⋯ , 𝑙_1+∑ _𝛼

𝑖 𝐶−1

𝑖=1 , ⋯ , 𝑙𝐷}

Make Cartesian product of professor set and lecture set, we get

M = P × L = {(𝑝₁, 𝑙₁), (𝑝₁, 𝑙₂), ⋯ , (𝑝₁, 𝑙_𝛼1), (𝑝₂, 𝑙_𝛼₁₊₁), ⋯ , (𝑝_𝐶, 𝑙_𝐷)} |M|=|L|=D

4.1.5 Class

Class set: C = {𝑐₁, 𝑐₂, ⋯ , 𝑐_𝑒, ⋯ 𝑐_𝐸}

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Each class has a certain number of students, defined by |𝑐_𝑒 . Such as

class 4 exactly have |𝑐₄ students.

4.2 Constraints

In a real case, each element is corresponding to others. We separate these relationships into two parts: “hard constraints” and “soft constraints”. Hard constraint means the situation that we must be exactly satisfied, otherwise it will be against the logic or the fact. Soft constraint is just our expectation and can be used to evaluate the model. In this model, they are defined as follows.

4.2.1 Hard constraints

1. H1: All allocated rooms are large enough.

2. H2: At the same time, each class cannot have a lecture in more than one room.

3. H3: No omission of classes in the timetable is demanded.

4. H4: At the same time, each professor only can teach one lecture in one classroom.

5. H5: At the same time, every room cannot be arranged for more than one lecture.

4.2.2 Soft constraints

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to the seating capacity of the allocated room.

2. S2: Students’ preference of 4 time-plots of one day is different. For example, the best time to study is in the morning. Therefore we should arrange lectures in the good memory time-plot as well as possible.

3. S3: The proper rest time is necessary. It is better to scatter the lectures of each class.

4.3 Solving procedures

Now we solve this problem by GA by following the steps introduced before. The first step which is very important is initializing code. A matrix is a much better coding type to be chosen than the linear type. And the calculation will be simplified if we use binary numbers as the elements of matrix. Since we have 5 elements, it is very hard to show how the 5 hard conditions are satisfied only in one matrix. So the author uses two

matrices (one is C~N, the other is N~M) to analyze the hard constraints

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4.3.1 Creating initialization code based on 5 hard-constraints

The Matrix 𝐶~𝑁 = [𝑎_𝑖,𝑗]_𝐸×20𝐵 is given in Figure 4.3.1-1.

(𝒕_𝟏, 𝒓_𝟏) (𝒕_𝟏, 𝒓_𝟐) ⋯ (𝒕_𝒂, 𝒓_𝒃) ⋯ (𝒕_𝟐𝟎, 𝒓_𝑩) 𝒄_𝟏 𝒄_𝟐 ⋯ 𝒄_𝒊 𝑎_𝑖,𝑗 ⋯ 𝒄_𝑬 Figure 4.3.1-1 Matrix 𝐶~𝑁 = [𝑎_𝑖,𝑗]_𝐸×20𝐵 𝑎_𝑖,𝑗 = {_{0 𝑖𝑓 𝑒𝑣𝑒𝑛𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑝𝑝𝑒𝑛}1 𝑖𝑓 𝑒𝑣𝑒𝑛𝑡 ℎ𝑎𝑝𝑝𝑒𝑛𝑠

where the event means that the class 𝑖 will have a lecture in room 𝑟_𝑏

at time 𝑡_𝑎.

H1: All allocated rooms are large enough.

If the capacity of the room 𝑘 is smaller than the number of the students of the class 𝑖, this room 𝑘 should not be arranged to this class 𝑖. Thus

the event will not happen, shown by 𝑎_𝑖,𝑘 = 0 no matter in which

time-plot. Specifically, we have 20 time-plots, in another words, 20

columns which include room 𝑘 will be found in the matrix C~N.

According to the room set, we have B rooms in total. Thus in every

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which is indicated by 𝑎_{𝑖,𝑘+𝐵∙𝑚} = 0, (𝑚 = 0, ⋯ ,19) . The sum of 𝑎-values of all of 20 time-plots will be equal to 0 as well. Therefore H1 can be defined by Function 4.3-1.

Function 4.3-1 If |𝑐_𝑖|>|𝑟_𝑘|, then ∑ 𝑎_{𝑖,𝑘+𝐵∙𝑚} 19 𝑚=0 = 0 Example 4.3.1-1:

If |𝑐₃|>|𝑟₅|, then in matrix C~N, cells 𝑎3,5, 𝑎3,𝐵+5, ⋯, 𝑎3,5+19𝐵 in row

𝑐₃ take the zero values like in Figure 4.3.1-2.

⋯ (𝒕_𝟏, 𝒓_𝟓) ⋯ (𝒕_𝟐, 𝒓_𝟓) ⋯ (𝒕_𝟏𝟗, 𝒓_𝟓) ⋯ (𝒕_𝟐𝟎, 𝒓_𝟓) ⋯

𝒄_𝟑 ⋯ 𝑎3,5=0 ⋯ 𝑎3,𝐵+5=0 ⋯ 𝑎3,5+18𝐵=0 ⋯ 𝑎3,5+19𝐵=0 ⋯

Figure 4.3.1-2 An example showing H1 ∑ 𝑎_{3,5+𝐵∙𝑚}

19

𝑚=0

= 0

Furthermore, maybe in the same time, more than two classes have the same lecture in the same room. So the sum of the number of the students of the classes which have the same lecture should be less than or equal to the capacity of the arranged room.

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Function 4.3-2 ∑ 𝑎_𝑖,𝑗 ∙ 𝑐_𝑖 𝐸 𝑖=1 ≤ 𝑟 𝑗−[𝑗−1_{𝐵 ]∙𝐵} j = 1,2, ⋯ ,20B

[ ] is an integer less than, i.e., [5.4]=5

If 𝑎_𝑖,𝑗 = 1, it means the class 𝑖 have a lecture, so ∑𝐸_𝑖=1𝑎_𝑖,𝑗 ∙ 𝑐_𝑖

means the sum of the number of the students. We cannot say the

arranged room is room j, since j may be larger than B. We define the

column-group that contain 𝑡₁ from the column 1 to B is 𝑡₁ block, and

between B+1 to 2B is 𝑡₂ block, and etc., thus [𝑗−1_𝐵 ] means how many

time-slots blocks we have passed. Then use 𝑗 − [𝑗−1_𝐵 ] ∙ 𝐵 to obtain the

number of allocated room. (Here we cannot use [_𝐵𝑗]. If 1 ≤ 𝑗 ≤ 𝐵 (in

time-slot 𝑡₁), [_𝐵𝑗] should equal to 0, then 𝑗 − [𝑗−1_𝐵 ] ∙ 𝐵 = 𝑗. But when 𝑗 = 𝐵, [𝑗

𝐵] will get 1, 𝑗 − [

𝑗−1

𝐵 ] ∙ 𝐵 = 0, which is impossible, so we use

[𝑗−1 𝐵 ] instead [ 𝑗 𝐵], in case 𝑗 = 𝑘𝐵, (𝑘 ∈ 𝑁 +_).) Example 4.3.1-2:

Assume that we have 10 rooms in total (B=10), at time-plot 𝑡₆, class 𝑐₄,

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⋯ (𝒕_𝟔, 𝒓_𝟕) ⋯

𝒄_𝟒 ⋯ 𝑎_4,7+5𝐵 = 𝑎_4,57 = 1 ⋯

𝒄_𝟔 ⋯ 1 ⋯

𝒄_𝟗 ⋯ 1 ⋯

Figure 4.3.1-3 More than 2 classes’ situation in H1 So 𝑐₄ ∙ 1 + 𝑐₆ ∙ 1 + 𝑐₉ ∙ 1 ≤ |𝑟_7+5𝐵−[6+5𝐵 𝐵 ]∙𝐵| = |𝑟7+5∙10−[ 6+5∙10 10 ]∙10| = |𝑟_57−[56 10]∙10| = 𝑟57−50 = 𝑟7 .

H2: At the same time, every one class cannot have a lecture in more than one room. We depict H2 in Figure 4.3.1-4.

(𝒕_𝟏, 𝒓_𝟏) ⋯ (𝒕_𝟏, 𝒓_𝑩) (𝒕_𝟐, 𝒓_𝟏) ⋯ (𝒕_𝟐, 𝒓_𝑩_{) ⋯ (𝒕}

𝟏+[𝒋−𝟏_{𝑩 ]}, 𝒓𝒋−[𝒋−𝟏_{𝑩 ]∙𝑩}) ⋯

𝒄_𝒊 𝑎𝑖,1 𝑎𝑖,𝐵 𝑎𝑖,1+𝐵 𝑎𝑖,2𝐵 𝑎𝑖,𝑗 ⋯

If 𝑎_2,1 = 𝑎_2,𝐵 = 1, which means in the time 𝑡₁ class no.2 has a lecture

in room no.1 and has a lecture in room no. B as well, that is ridiculous. Therefore, for each class, the sum of 𝑎-values of each time-plot block cannot be larger than 1.

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Function 4.3-3 ∑ 𝑎_{𝑖,𝑗+𝐵∙𝑛} 𝐵 𝑗=1 ≤ 1 i = 1,2, ⋯ , E n = 0, ⋯ , 19

H3: No omission of classes in the timetable. We describe H3 in Figure 4.3.1-5.

(𝑡₁, 𝑟₁) (𝑡₁, 𝑟₂) ⋯ (𝑡₂₀, 𝑟_𝐵)

𝑐_𝒊

If the sum of row no. 𝑖 is 0, it means the class no. 𝑖 does not have any lectures, which is not allowed.

So for each class

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We reveal the Matrix 𝑁~𝑀 = [𝑏_𝑠,𝑡]_20𝐵×𝐷 in Figure 4.3.1-6. (𝒑𝟏, 𝒍𝟏) ⋯ (𝒑𝟏, 𝒍𝜶𝟏) (𝒑𝟐, 𝒍𝜶𝟏+𝟏) ⋯ (𝒑𝟐, 𝒍𝜶𝟏+𝜶𝟐) ⋯ (𝒑𝒄, 𝒍𝒅) ⋯ (𝒑𝑪, 𝒍𝑫) (𝒕𝟏, 𝒓𝟏) (𝒕𝟏, 𝒓𝟐) ⋯ (𝒕𝒂, 𝒓𝒃) 𝑏_𝑠,𝑡 ⋯ (𝒕_𝟐𝟎, 𝒓𝑩) Figure 4.3.1-6 Matrix 𝑁~𝑀 = [𝑏_𝑠,𝑡]_20𝐵×𝐷 𝑏_𝑠,𝑡 = {1 𝑖𝑓 𝑒𝑣𝑒𝑛𝑡 ℎ𝑎𝑝𝑝𝑒𝑛𝑠 _{0 𝑖𝑓 𝑒𝑣𝑒𝑛𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑝𝑝𝑒𝑛}

where event means that the professor 𝑝_𝑐 will teach lecture 𝑙_𝑑 in room

𝑟_𝑏 at time 𝑡_𝑎.

H4: At the same time, each professor only can teach one lecture in one classroom. H4 is performed in Figure 4.3.1-7 and Figure 4.3.1-8.

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(𝒑_𝟏, 𝒍_𝟏) ⋯ (𝒑_𝟏, 𝒍_𝜶_𝟏) (𝒑_𝟐, 𝒍_𝜶_𝟏_+𝟏) ⋯ (𝒑_𝟐, 𝒍_𝜶_𝟏_+𝜶_𝟐) ⋯ (𝒕_𝟏, 𝒓_𝟏) Σ ≤ 1 Σ ≤ 1 ⋯ ⋯ (𝒕_𝟏, 𝒓_𝑩) (𝒕_𝟐, 𝒓_𝟏) Σ ≤ 1 Σ ≤ 1 ⋯ ⋯ (𝒕_𝟐, 𝒓_𝑩) ⋯ ⋯ ⋯ ⋯

Figure 4.3.1-8 H4 constraint (in detail) At first, we introduce two definitions:

1) 𝑏_𝑠. means the sum of row s, shown as ∑𝐷_𝑡=1𝑏_𝑠,𝑡;

2) 𝑏_.𝑡 means the sum of column t, shown as ∑20𝐵_𝑠=1𝑏_𝑠,𝑡.

We use ∑𝐵_𝑠=1𝑏_{(𝑠+𝑓∙𝐵).}, 𝑓 = 0, 1, ⋯ , 19 to show the sum of time-plot

𝑡_𝑓+1 block. The first professor can teach 𝛼₁ kinds of courses (the

columns which include 𝑝₁ are from 1 to 𝛼₁), so the sum of the first

professor block is ∑𝛼_𝑡=11 𝑏_.𝑡. And the second professor can teach 𝛼₂

courses (from column 𝛼₁ + 1 to 𝛼₂), thus the sum is ∑𝛼_𝑡=12 𝑏_.(𝑡+𝛼₁).

Therefore ∑ 𝑏_.(𝑡+∑ℎ _𝛼_𝑔

𝑔=0 ), ℎ = 0,1, ⋯ , 𝐶 − 1

𝛼ℎ+1

𝑡=1 means the sum of

professor 𝑝_ℎ+1 block. Then we create the function of H4 constraint is as

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Function 4.3-5 ∑ ∑ 𝑏_{(𝑠+𝑓∙𝐵),(𝑡+∑}ℎ _𝛼_𝑔 𝑔=0 ) 𝛼ℎ+1 𝑡=1 𝐵 𝑠=1 ≤ 1 𝑓 = 0,1, ⋯ ,19, ℎ = 0,1, ⋯ , 𝐶 − 1, 𝛼₀ = 0 Example 4.3.1-3:

Professor 𝑝₄ only can teach one lecture at time 𝑡₂ in one room, which

is the sum of 𝑡₂− 𝑝₄ block should be less than or equal to 1.

In this case, we know that 𝑓 = 1, ℎ = 3,

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⋯ (𝒑_𝟒, 𝒍_𝟏+𝜶_𝟏_+𝜶_𝟐_+𝜶_𝟑) ⋯ (𝒑_𝟒, 𝒍_𝜶_𝟏_+𝜶_𝟐_+𝜶_𝟑_+𝜶_𝟒) ⋯ ⋯ ⋯ (𝒕_𝟏, 𝒓_𝑩) ⋯ (𝒕_𝟐, 𝒓_𝟏) ⋯ Σ ≤ 1 ⋯ (𝒕_𝟐, 𝒓_𝑩) ⋯ ⋯ ⋯ Figure 4.3.1-9 An example of H4

H5: At the same time, every room cannot be arranged for more than one lecture. H5 is depicted in Figure 4.3.1-10.

(𝒑_𝟏, 𝒍_𝟏) ⋯ (𝒑_𝟏, 𝒍_𝜶𝟏) ⋯ (𝒑_𝑪, 𝒍_𝑫) (𝒕_𝟏, 𝒓_𝟏)

If 𝑏₁₂ = 𝑏₁₆ = 1 which means lecture 2 and lecture 6 both are taught in

room 1 at time-plot 1, that is impossible. Therefore the sum of each row should be less than or equal to 1. We formulate H5 by Function 4.3-6.

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We have already satisfied all of hard constraints, and now we combine

C~N matrix and N~M matrix together, in order to obtain the final initial

chromosome code. The Matrix 𝐶~𝑁~𝑀 = [𝑐_𝑝,𝑞]_{20𝐵𝐸×𝐷} is shown in

Figure 4.3.1-11. (𝒑_𝟏, 𝒍_𝟏) ⋯ (𝒑_𝟏, 𝒍_𝜶𝟏) (𝒑_𝟐, 𝒍_{𝜶𝟏+𝟏}) ⋯ (𝒑_𝟐, 𝒍_{𝜶𝟏+𝜶𝟐}) ⋯ (𝒑_𝑪, 𝒍_𝑫) (𝒄_𝟏, 𝒕_𝟏, 𝒓_𝟏) (𝒄_𝟏, 𝒕_𝟏, 𝒓_𝟐) ⋯ 𝑐_𝑝,𝑞 (𝒄_𝑬, 𝒕_𝟐𝟎, 𝒓_𝑩) Figure 4.3.1-11 Matrix 𝐶~𝑁~𝑀 = [𝑐_𝑝,𝑞]_{20𝐵𝐸×𝐷}

when 𝑐₁₁ = 1, then it shows that professor 𝑝₁ will teach class 𝑐₁

lecture 𝑙₁ in room 𝑟₁ at time-plot 𝑡₁, which means in matrix C~N

𝑎₁₁ = 1 and in matrix N~M 𝑏₁₁ = 1. If one of 𝑎-value or b-value is 0,

the event will not happen, thus c value should be equal to 0.

The combining function is following:

Function 4.3-7 𝑐_𝑝,𝑞 = 𝑎_([𝑝−1 20𝐵]+1),(𝑝−[ 𝑝−1 20𝐵]∙20𝐵) ∙ 𝑏(𝑝−[ 𝑝−1 20𝐵]∙20𝐵),𝑞 Example 4.3.1-4:

For 𝑐_35,20 in the row 35 and column 20, we should find its position in

matrix C~N~M first and then connect it to matrix C~N and matrix N~M.

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is included from row 1 to row 20B, 𝑐₂ is belonging to the rows between

row 20B+1 and row 40B and so on. So use [_20𝐵34] + 1 to obtain the order

of class (𝑐_[34

20𝐵]+1), same as the number of the row in matrix C~N. From

𝑝 − [𝑝−1

20𝐵] ∙ 20𝐵, we obtain the order of time-room group, same as the

column in matrix N~M. The row of matrix N~M is in accordance with

the column of matrix C~N. And the columns of matrix N~M and matrix

C~N~M are totally the same. Now we know that 𝑐_35,20 is

corresponding to 𝑎_([34 20𝐵]+1),(35−[ 34 20𝐵]∙20𝐵) and 𝑏(35−[ 34 20𝐵]∙20𝐵),20, only when 𝑎_([34 20𝐵]+1),(35−[ 34 20𝐵]∙20𝐵) and 𝑏(35−[ 34

20𝐵]∙20𝐵),20 are both equal to 1,

as a result 𝑐35,20 = 1, otherwise 𝑐35,20 = 0.

Since if and only if 𝑎_([𝑝−1

20𝐵]+1),(𝑝−[ 𝑝−1

20𝐵]∙20𝐵) = 1 and 𝑏(𝑝−[ 𝑝−1

20𝐵]∙20𝐵),𝑞 = 1,

then 𝑐_𝑝,𝑞 = 1. Thus 𝑐_𝑝,𝑞 still satisfies H1, H2, H4, H5 which ask for the

upper bound. An auxiliary condition should be added in case when H3 is broken. Function 4.3-8 ∑ ∑ 𝑐_{𝑝+𝐾∙20𝐵} 20𝐵 𝑝=1 𝐷 𝑞=1 ≥ 1 𝐾 = 0, 1, ⋯ , 𝐸 − 1

So the sum of every class block in matrix C~N~M cannot be equal to 0,

to make sure that no omission of classes happens.

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Initialization of chromosomes

Procedure Initialization (for each chromosome)

Step1: generate matrix C~N and matrix N~M randomly, 𝑎𝑖,𝑗and 𝑏𝑠,𝑡

equals to 0 or 1

Step2: check H1, H2, H3 in matrix C~N, check H4, H5 in matrix N~M, if they are satisfied, go to step 3, else, go back to step1

Step3: combine matrix C~N and matrix N~M to be matrix C~N~M Figure 4.3.1-12 Initialization

Modify

From the combining function, we will know that if one of 𝑎-value and

𝑏-value is equal to 0 then the corresponding c-value 𝑐_𝑝,𝑞 = 0. But

actually if 𝑐_𝑝,𝑞 = 0 which means none of class will have lectures in that

room at that time, therefore the real 𝑎_([𝑝−1

20𝐵]+1),(𝑝−[ 𝑝−1

20𝐵]∙20𝐵) value and

𝑏_(𝑝−[𝑝−1

20𝐵]∙20𝐵),𝑞 value should both be equal to 0. For instance if 𝑎11 =

1(which means 𝑐₁ have a lecture in 𝑟₁ at 𝑡₁) and ∑𝐶_𝑡=1𝑏_1,𝑡 = 0(𝑏_1. =

0 which means none of professors will teach lectures in 𝑟₁ at 𝑡₁), after

calculation or observing the table we get ∑𝐶_𝑞=1𝑐_1,𝑞 = 0 (𝑐_1. = 0) which

means that 𝑐1 does not have any lectures in 𝑟1 at 𝑡1, so the real 𝑎11

is 0. Then we need to correct the matrix C~N and matrix N~M. In

accordance with the logic, if 𝑐_𝑝,𝑞 = 1, then

𝑎

([𝑝−1_{20𝐵]+1),(𝑝−[}𝑝−1_{20𝐵]∙20𝐵)} = 1 𝑎𝑛𝑑 𝑏(𝑝−[𝑝−1_{20𝐵]∙20𝐵),𝑞} = 1.

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𝑎_([𝑝−1

20𝐵]+1),(𝑝−[𝑝−120𝐵]∙20𝐵) = 0 𝑎𝑛𝑑 𝑏(𝑝−[𝑝−120𝐵]∙20𝐵),𝑞 = 0.

We use 𝑎_𝑖,𝑗′ to show the new 𝑎-value after modification, and define

𝑏_𝑠,𝑡′ _{as the new 𝑏-value after modification.}

After comparing with matrix C~N and matrix C~N~M, we try to find

the relation of 𝑎_𝑖,𝑗 and 𝑐_𝑝,𝑞, shown by Figure 4.3.1-13.

(𝒕_𝟏, 𝒓_𝟏) ⋯ 𝑵_𝒋 ⋯ (𝒕_𝟐𝟎, 𝒓_𝑩) 𝒄_𝟏 ⋯ 𝒄_𝒊 𝑎_𝑖,𝑗 𝒄_𝑬 (𝒑_𝟏, 𝒍_𝟏) ⋯ (𝒑_𝑪, 𝒍_𝑫) (𝒄_𝟏, 𝒕_𝟏, 𝒓_𝟏) ⋯ (𝒄_𝒊, 𝑵_𝒋) 𝑐𝑗+(𝑖−1)∙20𝐵,𝑞 ⋯ (𝒄_𝑬, 𝒕_𝟐𝟎, 𝒓_𝑩)

Figure 4.3.1-13 The relationship between 𝑎_𝑖,𝑗 and 𝑐_𝑝,𝑞

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Matrix C~N includes class, room and time-plot. These three elements

are the same as in the contents of each row in matrix C~N~M.

Therefore the 𝑎-value 𝑎_𝑖,𝑗 is corresponding to the sum of the whole

row (𝑐_𝑖, 𝑁_𝑗) of matrix C~N~M. We thus know that matrix C~N~M

satisfies H2 and H4, so ∑𝐷_𝑞=1𝑐_𝑝,𝑞 ≤ 1, 𝑝 = 1, 2, ⋯ , 20𝐵𝐸 to insure that

the new 𝑎_𝑖,𝑗′ is a binary number.

After modification Function 4.3-9 𝑎_𝑖,𝑗′ _{= ∑ 𝑐} 𝑗+(𝑖−1)∙20𝐵,𝑞 𝐷 𝑞=1

Then we modify the matrix N~M, the comparison between matrix N~M

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(𝒑_𝟏, 𝒍_𝟏) ⋯ 𝑴𝒕 (𝒑_𝑪, 𝒍_𝑫) (𝒕_𝟏, 𝒓_𝟏) ⋯ 𝐍_𝒔 𝑏_𝑠,𝑡 (𝒕_𝟐𝟎, 𝒓_𝑩) (𝒑_𝟏, 𝒍_𝟏) ⋯ 𝑴_𝒕 ⋯ (𝒑_𝑪, 𝒍_𝑫) ⋯ (𝒄_𝟏, 𝑵_𝑺) 𝑐_𝑠,𝑡 ⋯ (𝒄_𝟐, 𝑵_𝑺) 𝑐_{𝑠+20𝐵,𝑡} ⋯ ⋯ (𝒄_𝑬, 𝑵_𝑺) 𝑐_{𝑠+(𝐸−1)∙20𝐵,𝑡} ⋯

Figure 4.3.1-14 The relationship between 𝑏_𝑠,𝑡 and 𝑐_𝑝,𝑞

If the event will not happen at 𝑁_𝑆, then ∑𝐸−1_𝑛=0𝑐_{𝑠+20𝐵∙𝑛,𝑡} = 0 and

𝑏_𝑠,𝑡 = 0 as well.

If one class or more than one class have the same lecture at 𝑁_𝑆, then

∑𝐸−1𝑐_{𝑠+20𝐵∙𝑛,𝑡}

𝑛=0 ≥ 1. Since 𝑏𝑠,𝑡 is a binary number, we apply 𝜃 function.

After modification

𝜃(Σ) = 𝑏_𝑠,𝑡′

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Function 4.3-10 𝑏_𝑠,𝑡′ _{= 𝜃 (∑ 𝑐} 𝑠+20𝐵∙𝑛,𝑡 𝐸−1 𝑛=0 ) (𝜃(𝑥) = {0, 𝑥 ≤ 0 _{1, 𝑥 > 0)}

4.3.2 Formulating the fitness function according to 3 soft-constraints

The soft-constraints can help us evaluate the model.

S1: After making sure that the arranged room is large enough, we should think about the using rate of the room, of course the higher rate the better. Therefore the number of the class students had better is similar to the seating capacity of the allocated room. We classify the rooms into small-sized, middle-sized, and large-sized based on the different seating capacity. And do the same thing with the set of classes. It is very easy to notice that there is a little bit waste if we use a large-sized room for a small-sized class. Preference degree of utilization of rooms is given by Table 4.3.2-1.

Table 4.3.2-1 Preference degree of utilization of rooms

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Assume that classes 𝑐₁~𝑐_ℎare small-sized, 𝑐_ℎ+1~𝑐_𝑙 are middle-sized,

𝑐_𝑙+1~𝑐_𝐸 are large-sized, rooms 𝑟₁~𝑟_𝑛 are small-sized, 𝑟_𝑛+1~𝑟_𝑚 are

middle-sized, 𝑟_𝑚+1~𝑟_𝐵 are large-sized. And in each same size level, the

capacity of rooms should be larger than the number of students. We

create matrix 𝐷 = [𝑑_𝛼,𝛽]_𝐸×𝐵 shown in Figure 4.3.2-1to reveal S1.

𝒓_𝟏 ⋯ 𝒓_𝒏 𝒓_𝒏+𝟏 ⋯ 𝒓_𝒎 𝒓_𝒎+𝟏 ⋯ 𝒓_𝑩 𝒄_𝟏 10 6 2 ⋯ 𝒄_𝒉 𝒄_𝒉+𝟏 0 10 6 ⋯ 𝒄_𝒍 𝒄_𝒍+𝟏 0 0 10 ⋯ 𝒄_𝑬

Figure 4.3.2-1 S1 constraint (matrix 𝐷 = [𝑑_𝛼,𝛽]_𝐸×𝐵)

Small-sized Middle-sized Large-sized

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We use 𝑓₁ to show the sum of room used degrees of all classes. In S1, a class and a room are the only two elements that have been mentioned,

thus we just need matrix C~N to express 𝑓₁ in order to simplify the

calculation. Since 𝑎_𝑖,𝑗′ , 𝑗 is from 1 to 20B, but 𝑑_𝛼,𝛽, 𝛽 is from 1 to B, the

d-value should multiply the corresponding 𝑎-value of each time-plot.

Then get the sum. So Function 4.3-11 𝑓₁ = ∑ ∑ 𝑎_𝑖,𝑗′ ∙ 𝑑 𝑖,𝑗−[𝑗−1_{𝐵 ]∙𝐵} 𝐸 𝑖=1 20𝐵 𝑗=1

S2: The best time to study new knowledge is in the morning. And at noon, students are usually sleepy, could not focus on the study. The preference of students in different time-plots is given by Table 4.3.2-2.

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Table 4.3.2-2 Preference of timeslot

Timeslot 𝒕_𝟏 𝒕_𝟐 𝒕_𝟑 𝒕_𝟒 𝒕_𝟓 ⋯ 𝒕_𝟐𝟎

Expected value

12 8 4 6 12 6

We use 𝑓2 to define the sum of timeslot preference degrees of arranged

lectures. Only class element and time-plot element are shown in S2, so

we still use matrix C~N to calculate 𝑓₂. The preference degree of

time-plot 𝑡_1+4𝑛, 𝑛 = 0, 1, ⋯ ,4 is equal to 12, and the preference

degree of 𝑡_2+4𝑛 𝑖𝑠 8 and so on. Then the sum is

Function 4.3-12 𝑓₂ = ∑ ∑ ∑(12 ∙ 𝑎_{𝑖,𝑗+4𝐵∙𝑛}′ _{+ 8 ∙ 𝑎} 𝑖,𝑗+𝐵+4𝐵∙𝑛 ′ _{+ 4 ∙ 𝑎} 𝑖,𝑗+2𝐵+4𝐵∙𝑛′ 4 𝑛=0 𝐵 𝑗=1 𝐸 𝑖=1 + 6𝑎_{𝑖,𝑗+3𝐵+4𝐵∙𝑛}′ ₎

S3: The lectures of each class should be scattered as best as we can. The time between two lectures cannot be too long or too short. On one side, it is hard to remember the knowledge after long time. And on the other side, if the students have next lecture immediately, usually they will not get it very well.

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Table 4.3.2-3 Preference of time length degree of courses (Zhang & Xing, 2002) Time length between two adjacent lessons 𝒙 1,19 2,3,17,18 4,5,15,16 6,7,13,14 8,9,10,11,12 Expected value 𝐠(𝒙) 0 2 4 10 16 g(𝑥) = { 0, 𝑖𝑓 𝑥 = 1 or 19 2, 𝑖𝑓 𝑥 = 1,2,17 𝑜𝑟 18 4, 𝑖𝑓 𝑥 = 4,5,15 𝑜𝑟 16 10, 𝑖𝑓 𝑥 = 6,7,13 𝑜𝑟 14 16, 𝑖𝑓 𝑥 = 8,9,10,11 𝑜𝑟 12}

𝑥 means the time length between two adjacent lessons, if class 𝑐_𝑣

have lectures at time-plot 𝑡_𝑚 and 𝑡_𝑛, then the time length is

𝑥 = 𝑚 − 𝑛 . For example if 𝑐₁ has a lecture at 8:00-10:00 on Monday

(𝑡₁), and the next lecture is from 10:00 to 12:00 on Monday (𝑡₂), the

time length between these two lectures is 1, then according to Table 4.3.2-3 the expected value is 0.

In the row 𝑖 of matrix [𝑎_𝑖𝑗′ ]

𝐸×20𝐵 (the new matrix C~N after

modification, refer to Figure 4.3.1-1), find the positions of 𝑎-values

which are equal to 1 (such as Figure 4.3.2-2), which means the class 𝑐_𝑖

has lectures at those time-plots. We define the column of the first

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the time-plot of 𝑎_𝑖,𝛽′ ₁ is [𝛽1_𝐵−1] + 1, and 𝑎_𝑖,𝛽′ ₂ belongs to the time-plot [𝛽2−1

𝐵 ] + 1. So that class 𝑐𝑖 has a lecture at time-plot 𝑡[𝛽1−1_𝐵 ]+1 and the

next lecture is at 𝑡_[𝛽2−1 𝐵 ]+1

, thus the time length preference degree of these two lectures is g ([𝛽2_𝐵−1] − [𝛽1_𝐵−1]). If 𝑐_𝑖 has 𝑊 lectures every

week, so 𝑎_𝑖,𝛽′ _𝜔 = 1, 𝜔 = 1, ⋯ , 𝑊, (as in Figure 4.3.2-2) the time length

preference degree of 𝑐_𝑖 every week is

∑ g ([𝛽𝜔+1 − 1 𝐵 ] − [ 𝛽_𝜔 − 1 𝐵 ]) 𝑊−1 𝜔=1 ⋯ (𝒕_𝟒, 𝒓_𝟐) ⋯ (𝒕_𝟖, 𝒓_𝟔) ⋯ (𝒕_𝟏𝟓, 𝒓_𝟑) ⋯ 𝒄_𝒊 0 1 0 1 0 1 0

Figure 4.3.2-2 Find the positions of 𝑎_𝑖,𝛽′ _𝜔 = 1

Example 4.3.2-1:

Assume that class 𝑐₃ has 3 lectures every week, and we have 10 rooms

in total. Lecture 1 is at (𝑡₁, 𝑟₂), lecture 3 is at (𝑡₅, 𝑟₆), and lecture 1 is at

(𝑡₁₇, 𝑟₃). The schedule table is as in Figure 4.3.2-3,

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So the time length preference degree of class 𝑐₃ is ∑ g ([𝛽𝜔+1 − 1 𝐵 ] − [ 𝛽_𝜔 − 1 𝐵 ]) 𝑊−1 𝜔=1 = ∑ g ([𝛽𝜔+1− 1 𝐵 ] − [ 𝛽_𝜔 − 1 𝐵 ]) 2 𝜔=1 = g ([𝛽2− 1 𝐵 ] − [ 𝛽₁− 1 𝐵 ]) + g ([ 𝛽₃ − 1 𝐵 ] − [ 𝛽₂− 1 𝐵 ]) = g ([46 − 1 10 ] − [ 1 − 1 10 ]) + g ([ 163 − 1 10 ] − [ 46 − 1 10 ]) = g(4 − 0) + g(16 − 4) = g(4) + g(12) = 4 + 16 = 20

Define 𝑓₃ as the sum of time length preference degrees of all classes.

Function 4.3-13 𝑓₃ = ∑ ∑ g ([𝛽𝜔+1− 1 𝐵 ] − [ 𝛽_𝜔 − 1 𝐵 ]) 𝑊−1 𝜔=1 𝐸 𝑖=1

Then we obtain the fitness function as

𝑓 = 𝑓₁+ 𝑓₂ + 𝑓₃

Thus the procedure of fitness is performed as Figure 4.3.2-4. Evaluation of fitness of chromosomes

Procedure fitness (for each chromosome)

Stage1: modify matrix [𝑎𝑖,𝑗]_𝐸×20𝐵 based on matrix [𝑐𝑝,𝑞]_{20𝐵𝐸×𝐷} ,

become matrix [𝑎′_𝑖,𝑗]_𝐸×20𝐵 Stage2: calculate fitness function f

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4.3.3 Solving TTP by GA

Selection

Then we use roulette-wheel method (introduced in chapter 3) to select parents’ chromosomes.

Crossover

Here we declare 𝑃_{𝐶𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟} = 0.75, then we draw a random probability

𝑃_𝑐 ∈ [0, 1] for every pair of selected chromosomes. If 𝑃_𝑐 ≤ 𝑃_{𝐶𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟},

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After crossover of parents 𝑃𝑎𝑟𝑒𝑛𝑡₁ and 𝑃𝑎𝑟𝑒𝑛𝑡₂ , we get the

offspring’s chromosomes 𝑂₁ and 𝑂₂.

Mutation

Here we define 𝑃_𝑀 = 0.02, generate a random 𝑃_𝑚 ∈ [0,1] for each

element of each offspring’s chromosome. If 𝑃_𝑚 ≤ 𝑃_𝑀, then the 𝑐 value

will be changed to opposite figure, like:

𝑐_𝑝,𝑞: 0 → 1 𝑜𝑟 1 → 0

Obtain new generations

Then we will obtain the new generations. If they satisfy 5 hard constraints then go on, check the stopping criterion. Otherwise, choose new rows and columns to do crossover and mutation again until the offspring’s chromosomes are feasible solutions.

Stopping criterion

Stop after 3000 iterations, to see the result.

4.4 Result

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Table 4.4-1 The parameter list Parameters Value Number of time-slots 20 Number of rooms 5 Number of professors 10 Number of classes 4

Number of lectures of each professor 4

Crossover rate 0.75

Mutation rate 0.02

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We plot 3D graph in Figure 4.4-1.

Fitn

ess

Iteration

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If we take out the largest fitness value of every iteration population and draw a regression line based on these largest values, then they are performed like in Figure 4.4-2.

As expected, the results will be converged and we obtain the best result which is around 650.

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In order to check the effect of population on the result, we test different population sizes of 20, 30 and 40 with crossover rate 0.75, the result is seen in Figure 4.4-3.

Figure 4.4-3 Comparison with different population sizes

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And if we compare population sizes of 30 with different crossover rates: 0.70, 0.75 and 0.80, we will get the result in Figure 4.4-4.

Figure 4.4-4 Comparison with different crossover rates

As we can see, the result did not change so much with different crossover rates.

By comparing the time consumption with different population sizes as in Table 4.4-2, we find that the whole procedure with larger population size spends longer time to find the best result. And the figure of population-time (plotted in Figure 4.4-5) almost looks like a straight line.

P_𝐶 = 0.70

P_𝐶 = 0.75

P_𝐶 = 0.80

Fitness

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Table 4.4-2 The result of comparison Population Time[Sec] 10 43.1193 20 88.9606 30 133.7512 40 179.7556 50 228.0954 60 269.2885 70 313.9937 80 362.2176 90 426.1943 100 520.9020

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5 Conclusion and further work

This dissertation aims to introduce a very useful optimization method— the genetic algorithm. Besides what is mentioned in this dissertation, there are many extensions, such as multiple selection methods, complicated crossover ways and so on. When we face different practical problems, a suitable selection method should be chosen and instead of one crossover point (which is shown in this dissertation) two or more crossover point can also be used. Furthermore, when depending on the requirement of the question we could adjust mutation rate, for instance, as the addition of the iteration mutation rate becomes smaller to decrease mutation.

The genetic algorithm is particularly appropriate for solving timetabling and scheduling problems, and many scheduling software packages are based on this method. Other application domains of GA include computer-automated design, video game design, test of fuzzy control system, artificial intelligence design, mechatronics design and so on. The application based on GA to make a university time schedule is shown in this dissertation for the sake of popularizing the idea of GA better. Different from other dissertations which solved TTP by GA as well, the author creates a new model with considering that:

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2) More than one class will have a same lecture at the same time in the same room,

3) The best time to study is in the morning, and it is very difficult to remember the knowledge in the time-plot 13:00-15:00 since the students are sleepy.

Furthermore the binary matrix is adopted to create initialization code, which is the most effective. The author formulates this new model including 5 hard constraints and 3 soft constraints in a fresh way, so that the solving procedure is more intuitional and easier to be understandable.

And further investigation is necessary regarding the following issues: 1. Rooms can be classified into lecture rooms and labs.

2. Compulsory courses should be taken into consideration. According to the real requirement we compel the corresponding c-value to be 1. 3. In order to check the effectiveness of GA, a comparison with other

algorithms seems necessary.

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References

Chu, P., & Beasley, J. (1995). “A genetic algorithm for the generalized assignment problem”. European Journal of Operational Research.

Davis, L. (1991). “Handbook of Genetic Algorithms”. Van Nostrand Reinhold.

Wikipedia. (2010, 12). Retrieved 12 2011, from WIKIPEDIA: http://en.wikipedia.org/wiki/Genetic_algorithm

John H. Holland (1975). “Adaptation in Natural and Artificial Systems”. A Bradford Book.

Markoff, J. (1990-08-29). “What's the Best Answer? It is Survival of the Fittest”. New York Times.

Rutkowski, L. (2005). “Computational intelligence”. Polish Scientific Publishers PWN.

Obitko, M. (1998, 8). Retrieved 3 2012, from

http://www.obitko.com/tutorials/genetic-algorithms/selection.php

Cole, A. (1964, 7). “The preparation of examination timetables using a small store computer”. Computer Journal, pp. 117-121.

Wood, D. (1968, 11). “A system for computing university examination timetable”. Computer Journal, pp. 41-47.

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Table 4.3.2-1 Preference degree of utilization of rooms Table 4.3.2-2 Preference of timeslot

Table 4.3.2-3 Preference of time length degree of courses Table 4.4-1 The parameter list

Table 4.4-2 The result of comparison Figure 3.2.1-1 Solving procedures of GA

Figure 4.3.1-1 Matrix 𝐶~𝑁 = [𝑎_𝑖,𝑗]_𝐸×20𝐵

Figure 4.3.1-2 An example showing H1

Figure 4.3.1-3 More than 2 classes’ situation in H1 Figure 4.3.1-4 H2 constraint

Figure 4.3.1-6 Matrix 𝑁~𝑀 = [𝑏_𝑠,𝑡]_20𝐵×𝐷

Figure 4.3.1-8 H4 constraint (in detail) Figure 4.3.1-9 An example of H4 Figure 4.3.1-10 H5 constraint

Figure 4.3.1-11 Matrix 𝐶~𝑁~𝑀 = [𝑐_𝑝,𝑞]_{20𝐵𝐸×𝐷}

Figure 4.3.1-12 Initialization

Figure 4.3.1-13 The relationship between 𝑎_𝑖,𝑗 and 𝑐_𝑝,𝑞

Figure 4.3.1-14 The relationship between 𝑏_𝑠,𝑡 and 𝑐_𝑝,𝑞

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Figure 4.3.2-2 Find the positions of 𝑎_𝑖,𝛽′ _𝜔 = 1 Figure 4.3.2-3 An example of S3

Figure 4.3.2-4 Fitness Figure 4.3.3-1 Crossover

Description and Application of Genetic Algorithm